Heat and Specific Heat

Heat transfer, specific heat capacity, and calorimetry

🔥 Heat and Specific Heat

Heat vs. Temperature

Temperature: Measure of average kinetic energy of particles Heat (Q): Energy transferred between objects due to temperature difference

💡 Key Distinction: Temperature is a property; heat is energy in transit.

Heat flows spontaneously from hot to cold until thermal equilibrium is reached.

Units of Heat

Joule (J): SI unit of energy
Calorie (cal): Energy to raise 1 g of water by 1°C
- 1 cal = 4.186 J
Kilocalorie (kcal or Cal): Food "Calorie"
- 1 kcal = 1000 cal = 4186 J

Specific Heat Capacity

Specific heat (c) is the energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).

$Q = mc\Delta T$

where:

$Q$ = heat transferred (J)
$m$ = mass (kg)
$c$ = specific heat (J/kg·°C or J/kg·K)
$\Delta T$ = temperature change (°C or K)

Common Specific Heats:

| Substance | c (J/kg·°C) | |-----------|-------------| | Water | 4186 | | Ice | 2090 | | Steam | 2010 | | Aluminum | 900 | | Steel | 470 | | Copper | 387 | | Lead | 128 |

Water has the highest specific heat of common substances! This is why:

Coastal areas have moderate climates
Water is used in cooling systems
Land heats/cools faster than oceans

Sign Convention

$Q > 0$ : Heat absorbed (temperature increases)
$Q < 0$ : Heat released (temperature decreases)

$\Delta T = T_{final} - T_{initial}$

If $\Delta T > 0$ : temperature increases → heat absorbed If $\Delta T < 0$ : temperature decreases → heat released

Calorimetry

Calorimetry measures heat transfer using conservation of energy.

Principle: In an isolated system, heat lost = heat gained

$Q_{lost} + Q_{gained} = 0$

$\sum Q_i = 0$

Typical Setup:

Hot object placed in cold water
System isolated (insulated calorimeter)
Final temperature measured
Energy conservation: $Q_{hot} + Q_{cold} = 0$

Heat Capacity

Heat capacity (C) is heat needed to raise an object's temperature by 1°C:

$C = mc$

$Q = C\Delta T$

Units: J/°C

Difference:

Specific heat (c): Property of material (per kg)
Heat capacity (C): Property of specific object (total)

Methods of Heat Transfer

1. Conduction

Heat transfer through direct contact
Molecular collisions
Requires material medium
Example: Metal spoon in hot soup

2. Convection

Heat transfer by fluid motion
Hot fluid rises, cold sinks
Requires fluid (liquid or gas)
Example: Boiling water, ocean currents

3. Radiation

Heat transfer by electromagnetic waves
No medium required (works in vacuum)
All objects emit thermal radiation
Example: Sun warming Earth, heat lamps

Problem-Solving Strategy

Identify all objects exchanging heat
Set up energy conservation: $\sum Q = 0$
Write Q for each object: $Q = mc\Delta T$
Determine signs:
- Heating: $Q > 0$ , $\Delta T > 0$
- Cooling: $Q < 0$ , $\Delta T < 0$
Solve for unknown (usually final temperature or specific heat)
Check reasonableness: Final T should be between initial temperatures

Common Mistakes

❌ Using wrong sign for heat (lost vs. gained) ❌ Forgetting to convert units (g → kg, cal → J) ❌ Using Celsius in place of Kelvin inappropriately (ΔT is same, but not absolute T) ❌ Neglecting heat lost to surroundings (real calorimeters aren't perfectly insulated) ❌ Mixing up specific heat (c) and heat capacity (C)

📚 Practice Problems

1Problem 1easy

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? (c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Mass: $m = 2.0$ kg
Initial temp: $T_i = 20°\text{C}$
Final temp: $T_f = 80°\text{C}$
Specific heat: $c = 4186$ J/kg·°C

Find: Heat required $Q$

Solution:

Step 1: Calculate temperature change. $\Delta T = T_f - T_i = 80 - 20 = 60°\text{C}$

Step 2: Apply heat formula. $Q = mc\Delta T$ $Q = (2.0)(4186)(60)$ $Q = 502,320 \text{ J} = 502 \text{ kJ}$

Answer: 502 kJ of heat is required

This is equivalent to about 120 food Calories (kcal).

2Problem 2easy

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? (c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Mass: $m = 2.0$ kg
Initial temp: $T_i = 20°\text{C}$
Final temp: $T_f = 80°\text{C}$
Specific heat: $c = 4186$ J/kg·°C

Find: Heat required $Q$

Solution:

Step 1: Calculate temperature change. $\Delta T = T_f - T_i = 80 - 20 = 60°\text{C}$

Step 2: Apply heat formula. $Q = mc\Delta T$ $Q = (2.0)(4186)(60)$ $Q = 502,320 \text{ J} = 502 \text{ kJ}$

Answer: 502 kJ of heat is required

This is equivalent to about 120 food Calories (kcal).

3Problem 3medium

❓ Question:

A 200 g aluminum cup (c = 900 J/(kg·°C)) contains 500 g of water (c = 4186 J/(kg·°C)) at 20°C. A 100 g piece of copper (c = 387 J/(kg·°C)) at 80°C is placed in the water. Find the final equilibrium temperature.

💡 Show Solution

Solution:

Given:

Al cup: m₁ = 0.200 kg, c₁ = 900 J/(kg·°C), T₁ = 20°C
Water: m₂ = 0.500 kg, c₂ = 4186 J/(kg·°C), T₂ = 20°C
Cu: m₃ = 0.100 kg, c₃ = 387 J/(kg·°C), T₃ = 80°C

Energy conservation: Heat lost by copper = Heat gained by water + aluminum

m₃c₃(T₃ - T_f) = m₁c₁(T_f - T₁) + m₂c₂(T_f - T₂)

Since T₁ = T₂ = 20°C: (0.100)(387)(80 - T_f) = (0.200)(900)(T_f - 20) + (0.500)(4186)(T_f - 20)

3096 - 38.7T_f = 180T_f - 3600 + 2093T_f - 41,860 3096 - 38.7T_f = 2273T_f - 45,460 48,556 = 2311.7T_f T_f = 21.0°C

The large mass and high specific heat of water dominates.

4Problem 4medium

❓ Question:

💡 Show Solution

Solution:

Given:

Al cup: m₁ = 0.200 kg, c₁ = 900 J/(kg·°C), T₁ = 20°C
Water: m₂ = 0.500 kg, c₂ = 4186 J/(kg·°C), T₂ = 20°C
Cu: m₃ = 0.100 kg, c₃ = 387 J/(kg·°C), T₃ = 80°C

Energy conservation: Heat lost by copper = Heat gained by water + aluminum

m₃c₃(T₃ - T_f) = m₁c₁(T_f - T₁) + m₂c₂(T_f - T₂)

Since T₁ = T₂ = 20°C: (0.100)(387)(80 - T_f) = (0.200)(900)(T_f - 20) + (0.500)(4186)(T_f - 20)

3096 - 38.7T_f = 180T_f - 3600 + 2093T_f - 41,860 3096 - 38.7T_f = 2273T_f - 45,460 48,556 = 2311.7T_f T_f = 21.0°C

The large mass and high specific heat of water dominates.

5Problem 5medium

❓ Question:

A 0.50 kg piece of aluminum at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature? (c_Al = 900 J/kg·°C, c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Aluminum: $m_{Al} = 0.50$ kg, $T_{Al,i} = 100°\text{C}$ , $c_{Al} = 900$ J/kg·°C
Water: $m_w = 2.0$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C

Find: Final temperature $T_f$

Solution:

Step 1: Set up energy conservation. $Q_{Al} + Q_w = 0$ $m_{Al}c_{Al}\Delta T_{Al} + m_w c_w \Delta T_w = 0$

Step 2: Express temperature changes. $m_{Al}c_{Al}(T_f - T_{Al,i}) + m_w c_w(T_f - T_{w,i}) = 0$

Step 3: Substitute values. $(0.50)(900)(T_f - 100) + (2.0)(4186)(T_f - 20) = 0$ $450(T_f - 100) + 8372(T_f - 20) = 0$ $450T_f - 45,000 + 8372T_f - 167,440 = 0$ $8822T_f = 212,440$ $T_f = 24.1°\text{C}$

Verification:

Aluminum cools: $\Delta T_{Al} = 24.1 - 100 = -75.9°\text{C}$ ✓
Water warms: $\Delta T_w = 24.1 - 20 = 4.1°\text{C}$ ✓
Final T between initial temperatures ✓

Answer: Final temperature is 24.1°C

The water barely warms because it has much larger mass and specific heat!

6Problem 6medium

❓ Question:

A 0.50 kg piece of aluminum at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature? (c_Al = 900 J/kg·°C, c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Aluminum: $m_{Al} = 0.50$ kg, $T_{Al,i} = 100°\text{C}$ , $c_{Al} = 900$ J/kg·°C
Water: $m_w = 2.0$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C

Find: Final temperature $T_f$

Solution:

Step 1: Set up energy conservation. $Q_{Al} + Q_w = 0$ $m_{Al}c_{Al}\Delta T_{Al} + m_w c_w \Delta T_w = 0$

Step 2: Express temperature changes. $m_{Al}c_{Al}(T_f - T_{Al,i}) + m_w c_w(T_f - T_{w,i}) = 0$

Verification:

Aluminum cools: $\Delta T_{Al} = 24.1 - 100 = -75.9°\text{C}$ ✓
Water warms: $\Delta T_w = 24.1 - 20 = 4.1°\text{C}$ ✓
Final T between initial temperatures ✓

Answer: Final temperature is 24.1°C

The water barely warms because it has much larger mass and specific heat!

7Problem 7medium

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? Use c_water = 4186 J/(kg·°C).

💡 Show Solution

Solution:

Given: m = 2.0 kg, c = 4186 J/(kg·°C), ΔT = 80 - 20 = 60°C

Heat required: Q = mcΔT Q = (2.0)(4186)(60) Q = 5.02 × 10⁵ J or 502 kJ

8Problem 8hard

❓ Question:

A 0.20 kg piece of unknown metal at 150°C is placed in 0.50 kg of water at 20°C. The final temperature is 25°C. What is the specific heat of the metal? Assume no heat is lost to surroundings. (c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Metal: $m_m = 0.20$ kg, $T_{m,i} = 150°\text{C}$ , $c_m = ?$
Water: $m_w = 0.50$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C
Final: $T_f = 25°\text{C}$

Find: Specific heat of metal $c_m$

Solution:

Step 1: Set up energy conservation. $Q_m + Q_w = 0$ $m_m c_m \Delta T_m + m_w c_w \Delta T_w = 0$

Step 2: Calculate temperature changes. $\Delta T_m = 25 - 150 = -125°\text{C}$ $\Delta T_w = 25 - 20 = 5°\text{C}$

Step 3: Solve for $c_m$ . $m_m c_m \Delta T_m = -m_w c_w \Delta T_w$ $c_m = -\frac{m_w c_w \Delta T_w}{m_m \Delta T_m}$ $c_m = -\frac{(0.50)(4186)(5)}{(0.20)(-125)}$ $c_m = -\frac{10,465}{-25}$ $c_m = 419 \text{ J/kg·°C}$

Comparison with known metals:

Copper: 387 J/kg·°C
Steel: 470 J/kg·°C

Answer: Specific heat is 419 J/kg·°C

This is close to copper (387) or possibly a copper alloy.

9Problem 9hard

❓ Question:

💡 Show Solution

Given:

Metal: $m_m = 0.20$ kg, $T_{m,i} = 150°\text{C}$ , $c_m = ?$
Water: $m_w = 0.50$ kg, $T_{w,i} = 20°\text{C}$ , $c_w = 4186$ J/kg·°C
Final: $T_f = 25°\text{C}$

Find: Specific heat of metal $c_m$

Solution:

Step 1: Set up energy conservation. $Q_m + Q_w = 0$ $m_m c_m \Delta T_m + m_w c_w \Delta T_w = 0$

Step 2: Calculate temperature changes. $\Delta T_m = 25 - 150 = -125°\text{C}$ $\Delta T_w = 25 - 20 = 5°\text{C}$

Comparison with known metals:

Copper: 387 J/kg·°C
Steel: 470 J/kg·°C

Answer: Specific heat is 419 J/kg·°C

This is close to copper (387) or possibly a copper alloy.

10Problem 10medium

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? Use c_water = 4186 J/(kg·°C).

💡 Show Solution

Solution:

Given: m = 2.0 kg, c = 4186 J/(kg·°C), ΔT = 80 - 20 = 60°C

Heat required: Q = mcΔT Q = (2.0)(4186)(60) Q = 5.02 × 10⁵ J or 502 kJ

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics